Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $47.6$ years; the standard deviation is $8.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $64.4$ and $72.8$ years.
Explanation: $47.6$ $39.2$ $56$ $30.8$ $64.4$ $22.4$ $72.8$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $47.6$ years. We know the standard deviation is $8.4$ years, so one standard deviation below the mean is $39.2$ years and one standard deviation above the mean is $56$ years. Two standard deviations below the mean is $30.8$ years and two standard deviations above the mean is $64.4$ years. Three standard deviations below the mean is $22.4$ years and three standard deviations above the mean is $72.8$ years. We are interested in the probability of a bear living between $64.4$ and $72.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of bears between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular bear living between $64.4$ and $72.8$ years is $\color{orange}{2.35\%}$.